Question

Factors of $x^6-y^6$ is:

• A. $(x-y)(x+y)$
• B. $(x^2+y^2-xy)$
• C. $(x^2+y^2+xy)$
• D. All of the above

Answer 1

The correct option is D

All of the above

$x^6-y^6$ can be written as $\left(x^2\right){}^3-\left(y^2\right){}^3$. This is of the form $a^3-b^3$.
We know, $a^3-b^3$ = $(a-b)$$(a^2+ab+b^2)$ .
Here, $a$ = $x^2$ and $b$ = $y^2$ .
$\left(x^2\right){}^3-\left(y^2\right){}^3$ = $(x^2-y^2)$$\left(\right(x^2){}^2+(x^2\left)\right(y^2)+(y^2\left){}^2\right)$
Solving on R.H.S. side:-
$(x^2-y^2)$$(x^4+x^2{y}^2+y^4)$
Using Factor Theorem, split the middle term.
$(x^2-y^2)$$(x^4+2x^2{y}^2-x^2{y}^2+y^4)$
Rewriting,
$(x^2-y^2)$$(x^4+2x^2{y}^2+y^4-x^2{y}^2)$
$(x^2-y^2)$$\left(\right(x^2+y^2{)}^2-\left(xy{)}^2\right)$
We know, $a^2-b^2$ = $(a-b)$$(a+b)$ .
$(x^2-y^2)$$\left(\right(x^2+y^2{)}^2-\left(xy{)}^2\right)$ = $(x+y)(x-y)$$(x^2+y^2+xy)(x^2+y^2-xy)$